What is the early voltage of 2907a?

Discussion in 'Homework Help' started by Anna Hughes, Jul 26, 2016.

1. Anna Hughes Thread Starter New Member

Jul 26, 2016
6
0
hi! I was wondering if some one knew the early voltage for the 2907a pnp bjt or knew a circuit that I could use to find it. I have a power supply, multimeter, and function generator.

2. MrAl Well-Known Member

Jun 17, 2014
2,215
435
Hello there,

[Sorry i made a mistake in the original post that i corrected now]

You can probably find it in a spice model for the device.
Do you have a spice model?
If not you can download one or we could find one for you.
Look up what that parameter is called in spice then look at the spice model. I think forward is VAF or just VA.
If you want to measure, it is the point where several Ic vs Vce slopes (beyond the knee) extrapolate back to the very same point on the Vce axis.
In other words, if you measure the slope of the Ic vs Vce curve beyond the knee, then calculate 'b' in:
y=m*x+b

then set y=0, then solve for 'x', and this new 'x' is the Early voltage.

In the above,
y is the collector current at a point beyond the knee,
x is the collector emitter voltage at the same point beyond the knee to start with, then when the new x is solved for that is the Early voltage,
m is the slope you measured,
b calculated from the x,y point

Solving for b with measured slope m and measured point x1,y1 we get:
y1=m*x1+b
b=y-m*x

and in terms of the actual point (Vce,Ic) on the curve:
b1=Ic-m*Vce

Next, set y=0 to calculate the new 'x' which is the Early voltage:
y=m*x+b1
0=m*x+b1
m*x=-b1
x=-b1/m

x here is the Early voltage, or the negative of the Early voltage.

Simplified Example:
You measure ic1=1.01, ic2=1.02, Vce1=1, Vce2=2 (these two points must be measured to the right of the knee in the characteristic),
First calculate m:
m=(ic2-ic1)/(Vce2-Vce1)=0.01
Then calculate b:
b=y-m*x=ic1-m*Vce1=1.01-0.01*1=1.00
Finally, calculate the new x:
x=-b/m=-1.00/0.01=-100=Early voltage (or the negative of that could also be called the Early voltage)

[This looks correct now]

As a final note, some models use the collector base voltage rather than the collector emitter voltage as shown here.

Last edited: Jul 27, 2016
Anna Hughes likes this.

Jul 26, 2016
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4. SLK001 Active Member

Nov 29, 2011
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The Forward Early Voltage for 2N2907AJTXV by MOTOROLA is:

VAF = 17.63

(from the PSpice bipolar.lib).

5. Anna Hughes Thread Starter New Member

Jul 26, 2016
6
0
thanks so much!

6. JoeJester AAC Fanatic!

Apr 26, 2005
3,317
1,109
You know you can set it up and measure it like Mr Al explained

In my spice library the VAF is 56.7V and the VAR (reverse) is 28.3V. The measurements -54.5 to -54.6, depending on whether I used a constant current of 50 microamperes or 100 microamperes

Last edited: Jul 27, 2016
7. Bordodynov Active Member

May 20, 2015
601
175
See

Anna Hughes likes this.
8. Anna Hughes Thread Starter New Member

Jul 26, 2016
6
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This is fantastic! Thanks so much!

So you ran the simulation and it gave you Va = 122? So I can assume Va = 122?

Thanks so much!

9. Bordodynov Active Member

May 20, 2015
601
175
The of the transistor model parameter voltage Earley given VAF (Va forward). Therefore Yes Va=122 (in model VAF=120). This voltage is spread especially for different manufacturers. The accuracy of determining this parameter to 5% is sufficient.

10. Anna Hughes Thread Starter New Member

Jul 26, 2016
6
0
So I am going to assume that Va for the 2907 is 122. Does this Va change? Because there have been other posts saying the Va is 50 or 16.

11. WBahn Moderator

Mar 31, 2012
17,454
4,701
The Early voltage of most BJT transistors is not well controlled and most SPICE models use a convenient value that is hopefully in a useful ballpark range. Most transistor data sheets, particularly for non-RF transistors, don't give you enough information to even estimate it. I'm not sure I've ever seen a data sheet that even showed the classic family of Ic vs Vce curves that you see in virtually every text book.

I suspect that the reason might be that either the output resistance of the transistor is good enough or it isn't. If it is, then you don't worry about what it happens to be and if it isn't, then you use a different circuit topology that makes it good enough so that you don't worry about what it happens to be.

A lot of people just assume that Va = 100 V and, if it's an issue, proceed to measure it on real devices under conditions relevant to their application.

12. Bordodynov Active Member

May 20, 2015
601
175
See
.MODEL 2N2907S PNP(IS=2.32E-13 ISE=6.69E-16 ISC=1.65E-13 XTI=3.00 BF=3.08E2 BR=2.18E1 IKF=8.42E-1 IKR=1.00 XTB=1.5 VAF=1.41E2 VAR=1.35E1 VJE=3.49E-1 VJC=3.00E-1 RE=1.00E-2 RC=8.46E-1 RB=4.02E1 RBM=1.00E-2 IRB=1.25E-2 CJE=2.66E-11 CJC=1.93E-11 .00 FC=5.00E-1 NF=1.04 NR=1.12 NE=1.09 NC=1.13 MJE=4.60E-1 MJC=4.65E-1 TF=4.95E-10 TR=0 ITF=3.36E-1 VTF=6.54 XTF=1.87E1 EG=1.11 VCEO=80 ICRATING=5 MFG=SIEMENS)

.MODEL 2N2907P PNP(IS=650.6E-18 ISE=54.81f ISC=0 XTI=3 BF=231.7 BR=3.563 IKF=1.079 IKR=0 XTB=1.5 VAF=115.7 VAR=35 VJE=0.65 VJC=0.65 RE=0.15 RC=0.715 RB=10 CJE=19.82p CJC=14.76p XCJC=0.75 FC=0.5 NF=1 NR=1 NE=1.829 NC=2 MJE=0.3357 MJC=0.5383 TF=603.7p TR=111.3n ITF=0.65 VTF=5 XTF=1.7 EG=1.11 VCEO=40 ICRATING=600M MFG=NSC-PHILIPS)

.model 2N2907 PNP(IS=1E-14 VAF=120 BF=250 IKF=0.3 XTB=1.5 BR=3 CJC=8E-12 CJE=30E-12 TR=100E-9 TF=400E-12 ITF=1 VTF=2 XTF=3 RB=10 RC=.3 RE=.2 Vceo=40 Icrating=600m mfg=NXP)

.MODEL 2N2907A PNP(IS=3.81E-13 BF=154 VAF=139 IKF=0.14 ISE=1.53E-11
+ NE=2 BR=4 VAR=20 IKR=0.21 RB=2.21 RE=0.552 RC=0.221 CJE=1.56E-11 VJE=0.75
+ TF=6.36E-10 CJC=2.08E-11 VJC=0.75 TR=6.37E-8 VJS=0.75 XTB=1.5 )
.model Q2N2907 pnp ( IS=3.58E-14 VAF=26.4 BF=300 IKF=0.3416 NE=1.2861
+ ISE=3.830E-14 IKR=0.03 ISC=2.00E-12 NC=1.2 NR=1 BR=5 RC=0.4 CJC=1.80E-11
+ FC=0.5 MJC=0.45 VJC=0.8 CJE=2.65E-11 MJE=0.33 VJE=0.75 TF=4.10E-10
+ ITF=0.54 VTF=3 XTF=20 RE=1.4 TR=8.00E-08)

.MODEL QN2907 PNP IS=1.1E-12 BF=200 BR=6 NF=1.21
+ TF=5E-10 TR=34E-9
+ CJE=23PF VJE=.85 MJE=1.25 CJC=19PF VJC=.5 MJC=.2
+ RC=.6 IKF=100E-3 VAF=50 XTB=1.5 NE=1.9 ISE=1.3E-11

.MODEL 2N2907 PNP BF=217.797 BR=2.98475 CJC=36.6437p CJE=67.0252p IKF=1.12749
+ VAF=100 VJC=700m VJE=700m VTF=10 XTF=500m
+ MJE=595.493m NE=1.65086 NF=954.211m RC=405.548m RE=1.56766 TF=1n TR=29.1239n
+ IKR=1.00038 IS=10.0118f ISC=9.55131p ISE=1.01042p ITF=10m MJC=558.066m

13. Anna Hughes Thread Starter New Member

Jul 26, 2016
6
0
Great! Thanks!

14. JoeJester AAC Fanatic!

Apr 26, 2005
3,317
1,109
You had two examples on how to calculate the Early Voltage. That is what your professor wanted you to do, you know, show your work.

Some schools teach it's typically between 50 and 100. As you saw, different models gave different results. So, I wish you luck if you do this in the lab.

15. MrAl Well-Known Member

Jun 17, 2014
2,215
435
Hello there,

Just one caution here with any of these methods...
The transistor characteristic has to be taken beyond the knee or else the calculation can be way off.

For example, for a 2N2906 i can get VA=45 or VA=99 depending on how i bias it. The right value is 99 but i can only get that with the transistor biased correctly.
If it is biased too hard the slope is too steep so it crosses the x axis too soon leading to a number that is too low.

Also, if we calculate out the y=m*x+b technique we end up with this:
VA=(Ic2*Vce1-Ic1*Vce2)/(Ic2-Ic1)

where Vce1,Ic1 is one point on the curve, and Vce2,Ic2 is another point on the curve, provided again that the two points are both beyond the knee of the curve.

Code (Text):
1.
2.         p1        p2
3.   k ---O---------O--------------
4.   /
5.  /
6. /
7. --------------------------------------------------> Vce
8.
9. k is the knee of the Ic vs Vce curve
10.
11.

Last edited: Jul 29, 2016
16. Bordodynov Active Member

May 20, 2015
601
175
MrAl, I agree with you.
Here is a new calculation. The value VAF1 counted on your technique. I took Vbe = 0.7V.
See